What is the tangent of y=3x^3-x^2+2,(2,22) ?

The tangent of y=3x^3-x^2+2,(2,22) is y=32x-42

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tangent of y=3x^3-x^2+2,(2,22)

tangent of

y = 3 x^{3} - x^{2} + 2, (2, 22)

y = 32 x - 42

Solution steps

Find the slope of

y = 3 x^{3} - x^{2} + 2 : \frac{d y}{d x} = 9 x^{2} - 2 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 32

Find the line with slope m=

32

and passing through

(2, 22) : y = 32 x - 42

y = 32 x - 42

x \to - 1 lim (\frac{∣ x ∣ - 1}{x + 1})

h \to 0 lim (\frac{7 ( a + h ) ^{2} - 7 a ^{2}}{h})

f (x) = \frac{x ^{2}}{x ^{2} - 1}

\int \frac{1}{sin ( 5 x ) cos ( 5 x )} d x

\int 8 e^{- t} d t

What is the tangent of y=3x^3-x^2+2,(2,22) ?
The tangent of y=3x^3-x^2+2,(2,22) is y=32x-42